Check sibling questions

Example 47 - Find intervals in which f(x) = 3/10x4 - 4/5x3

Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 47 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Transcript

Example 47 Find intervals in which the function given by f(π‘₯) =3/10 π‘₯4 – 4/5 π‘₯^3– 3π‘₯2 + 36/5 π‘₯ + 11 is (a) strictly increasing (b) strictly decreasingf(π‘₯) = 3/10 π‘₯4 – 4/5 π‘₯^3– 3π‘₯2 + 36/5 π‘₯ + 11 Finding f’(𝒙) f’(π‘₯) = 3/10 Γ— 4π‘₯^3 – 4/5 Γ— 3π‘₯^2 – 3 Γ— 2x + 36/5 + 0 f’(π‘₯) = 12/10 π‘₯^3– 12/5 π‘₯^2– 6x + 36/5 f’(π‘₯) = 6/5 π‘₯^3βˆ’ 12/5 π‘₯^2– 6x + 36/5 f’(π‘₯) = 6(π‘₯^3/5βˆ’(2π‘₯^2)/5βˆ’π‘₯+6/5) f’(π‘₯) = 6((π‘₯^3 βˆ’ 2π‘₯^2βˆ’ 5π‘₯ + 6)/5) = 6/5 (π‘₯^3βˆ’2π‘₯^2βˆ’5π‘₯+6) = 6/5 (π‘₯βˆ’1)(π‘₯2βˆ’π‘₯βˆ’6) = 6/5 (π‘₯βˆ’1)(π‘₯2βˆ’3π‘₯+2π‘₯βˆ’6) = 6/5 (π‘₯βˆ’1)[π‘₯(π‘₯βˆ’3)+2(π‘₯βˆ’3)] = 6/5 (π‘₯βˆ’1)(π‘₯+2)(π‘₯βˆ’3) Hence, f’(𝒙) = πŸ”/πŸ“ (π’™βˆ’πŸ)(𝒙+𝟐)(π’™βˆ’πŸ‘) Putting f’(𝒙) = 0 πŸ”/πŸ“ (π’™βˆ’πŸ)(𝒙+𝟐)(π’™βˆ’πŸ‘) = 0 (π‘₯βˆ’1)(π‘₯+2)(π‘₯βˆ’3) = 0 Hence, x = –2 , 1 & 3 Plotting points on number line Hence, f(π‘₯) is strictly decreasing on the interval π‘₯ ∈ (βˆ’βˆž,βˆ’πŸ)& (𝟏 , πŸ‘) f(π‘₯) is strictly increasing on the interval π‘₯ ∈ (βˆ’πŸ,𝟏) & (πŸ‘ , ∞)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.